Class 12 Maths NCERT Solutions Chapter 7 Ex 7.5 – Integration by Partial Fractions | Boundless Maths
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Ex 7.5 Class 12 Maths NCERT Solutions

Class 12 Maths NCERT Solutions Chapter 7 Ex 7.5 – Integration by Partial Fractions

Chapter 7 Ex 7.5 covers integration by partial fractions — all 23 questions solved step by step, resolving rational functions into distinct linear, repeated linear, and irreducible quadratic pieces before integrating each one.

Every question here follows the same core routine: factor the denominator, write the fraction as a sum of simpler pieces with unknown constants, then solve for those constants by substituting convenient values of x or comparing coefficients. The exercise builds in difficulty by denominator type — distinct linear factors first (Q1–5), then repeated linear factors and irreducible quadratics (Q6–17), and finally denominators needing a substitution before decomposition, like eˣ − 1 or xⁿ + 1 (Q16, Q21). If the degree of the numerator is greater than or equal to the denominator, remember to divide first — skipping this step is the most common way this exercise catches students out. The two closing MCQs test the same decomposition skill under exam time pressure.

📐 Formula Cards 🤖 AI Question Bank
23Questions
Medium–HardDifficulty Mix
2026-27CBSE Syllabus

Chapter 7 Ex 7.5 Solutions — All 23 Questions

1

Integrate: \dfrac{x}{(x+1)(x+2)}

Easy +
Solution

Split into partial fractions: \dfrac{x}{(x+1)(x+2)} = \dfrac{A}{x+1}+\dfrac{B}{x+2}, so x=A(x+2)+B(x+1).

Putting x=-1: -1=A(1), so A=-1. Putting x=-2: -2=B(-1), so B=2.

Integrating: -\displaystyle\int\dfrac{dx}{x+1}+2\displaystyle\int\dfrac{dx}{x+2}=-\log|x+1|+2\log|x+2|.

\displaystyle\int \dfrac{x}{(x+1)(x+2)}\,dx = 2\log|x+2| - \log|x+1| + C
2

Integrate: \dfrac{1}{x^2-9}

Easy +
Solution

Direct application of the standard formula \displaystyle\int \dfrac{dx}{x^2-a^2} = \dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right| with a=3.

Using partial fractions: \dfrac{1}{(x-3)(x+3)}=\dfrac{A}{x-3}+\dfrac{B}{x+3}, so 1=A(x+3)+B(x-3). Putting x=3: 1=6A, so A=\dfrac16. Putting x=-3: 1=-6B, so B=-\dfrac16.

ShortcutThe numerator is 1 and the factors x-3 and x+3 differ by 6, so \dfrac{1}{x^2-9}=\dfrac16\left[\dfrac{1}{x-3}-\dfrac{1}{x+3}\right] directly: 1 over the smaller factor minus 1 over the bigger one, divided by the difference.

\displaystyle\int \dfrac{1}{x^2-9}\,dx = \dfrac16\log\left|\dfrac{x-3}{x+3}\right| + C
3

Integrate: \dfrac{3x-1}{(x-1)(x-2)(x-3)}

Medium +
Solution

Write \dfrac{3x-1}{(x-1)(x-2)(x-3)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}+\dfrac{C}{x-3}, so 3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2).

Putting x=1: 2=A(-1)(-2)=2A, so A=1. Putting x=2: 5=B(1)(-1), so B=-5. Putting x=3: 8=C(2)(1), so C=4.

Check: the numerator has degree 1, so there is no x^2 term on the left and A+B+C must be 0: 1-5+4=0. Integrating each term gives the answer below.

\displaystyle\int \dfrac{3x-1}{(x-1)(x-2)(x-3)}\,dx = \log|x-1| - 5\log|x-2| + 4\log|x-3| + C
4

Integrate: \dfrac{x}{(x-1)(x-2)(x-3)}

Medium +
Solution

Same decomposition pattern as Q3: x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2).

Putting x=1: 1=2A, so A=\dfrac12. Putting x=2: 2=-B, so B=-2. Putting x=3: 3=2C, so C=\dfrac32.

Check: A+B+C=\dfrac12-2+\dfrac32=0. Integrating each term gives the answer below.

\displaystyle\int \dfrac{x}{(x-1)(x-2)(x-3)}\,dx = \dfrac12\log|x-1| - 2\log|x-2| + \dfrac32\log|x-3| + C
5

Integrate: \dfrac{2x}{x^2+3x+2}

Easy +
Solution

Factor the denominator: x^2+3x+2=(x+1)(x+2). Then \dfrac{2x}{(x+1)(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}, so 2x=A(x+2)+B(x+1).

Putting x=-1: -2=A(1), so A=-2. Putting x=-2: -4=B(-1), so B=4.

Integrating: -2\log|x+1|+4\log|x+2|.

\displaystyle\int \dfrac{2x}{x^2+3x+2}\,dx = 4\log|x+2| - 2\log|x+1| + C
6

Integrate: \dfrac{1-x^2}{x(1-2x)}

Hard +
Solution

The numerator and denominator both have degree 2, so this is improper — divide first. Since x(1-2x)=x-2x^2, dividing 1-x^2 by x-2x^2 gives quotient \dfrac12, because (1-x^2)-\dfrac12(x-2x^2)=1-\dfrac{x}{2}. So

\dfrac{1-x^2}{x(1-2x)} = \dfrac12 + \dfrac{1-\frac{x}{2}}{x(1-2x)}, call the remainder term (1).

Decomposing (1) as \dfrac{A}{x}+\dfrac{B}{1-2x} gives 1-\dfrac{x}{2}=A(1-2x)+Bx. Putting x=0: A=1. Putting x=\dfrac12: \dfrac34=\dfrac{B}{2}, so B=\dfrac32.

Integrating: \displaystyle\int\left[\dfrac12+\dfrac1x+\dfrac{3/2}{1-2x}\right]dx=\dfrac{x}{2}+\log|x|+\dfrac32\cdot\left(-\dfrac12\right)\log|1-2x|.

\displaystyle\int \dfrac{1-x^2}{x(1-2x)}\,dx = \dfrac{x}{2} + \log|x| - \dfrac34\log|1-2x| + C
7

Integrate: \dfrac{x}{(x^2+1)(x-1)}

Medium +
Solution

Write \dfrac{x}{(x^2+1)(x-1)} = \dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}, so x=A(x^2+1)+(Bx+C)(x-1).

Putting x=1: 1=2A, so A=\dfrac12. Comparing the x^2 coefficients: 0=A+B, so B=-\dfrac12. Comparing the constant terms: 0=A-C, so C=\dfrac12.

So \dfrac{x}{(x^2+1)(x-1)}=\dfrac12\cdot\dfrac{1}{x-1}-\dfrac14\cdot\dfrac{2x}{x^2+1}+\dfrac12\cdot\dfrac{1}{x^2+1}. The middle term gives a log (its numerator is the derivative of the denominator) and the last term gives \tan^{-1}x.

\displaystyle\int \dfrac{x}{(x^2+1)(x-1)}\,dx = \dfrac12\log|x-1| - \dfrac14\log(x^2+1) + \dfrac12\tan^{-1}x + C
8

Integrate: \dfrac{x}{(x-1)^2(x+2)}

Medium +
Solution

Write \dfrac{x}{(x-1)^2(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+2}, so x=A(x-1)(x+2)+B(x+2)+C(x-1)^2.

Putting x=1: 1=3B, so B=\dfrac13. Putting x=-2: -2=9C, so C=-\dfrac29. Substitution cannot isolate A, so compare the x^2 coefficients: 0=A+C, so A=\dfrac29.

Integrating: \dfrac29\log|x-1|+\dfrac13\displaystyle\int(x-1)^{-2}dx-\dfrac29\log|x+2|, where \displaystyle\int(x-1)^{-2}dx=-\dfrac{1}{x-1}.

\displaystyle\int \dfrac{x}{(x-1)^2(x+2)}\,dx = \dfrac29\log|x-1| - \dfrac{1}{3(x-1)} - \dfrac29\log|x+2| + C
9

Integrate: \dfrac{3x+5}{x^3-x^2-x+1}

Medium +
Solution

Factor by grouping: x^3-x^2-x+1 = x^2(x-1)-(x-1) = (x-1)^2(x+1).

Write \dfrac{3x+5}{(x-1)^2(x+1)} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+1}, so 3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^2.

Putting x=1: 8=2B, so B=4. Putting x=-1: 2=4C, so C=\dfrac12. Comparing the x^2 coefficients: 0=A+C, so A=-\dfrac12.

Integrating: -\dfrac12\log|x-1|+4\displaystyle\int(x-1)^{-2}dx+\dfrac12\log|x+1|, where 4\displaystyle\int(x-1)^{-2}dx=-\dfrac{4}{x-1}.

\displaystyle\int \dfrac{3x+5}{x^3-x^2-x+1}\,dx = \dfrac12\log|x+1| - \dfrac12\log|x-1| - \dfrac{4}{x-1} + C
10

Integrate: \dfrac{2x-3}{(x^2-1)(2x+3)}

Hard +
Solution

Factor x^2-1=(x-1)(x+1), then write \dfrac{2x-3}{(x-1)(x+1)(2x+3)} = \dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{C}{2x+3}, so 2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1).

Putting x=1: -1=A(2)(5), so A=-\dfrac{1}{10}. Putting x=-1: -5=B(-2)(1), so B=\dfrac52. Putting x=-\dfrac32: -6=C\left(-\dfrac52\right)\left(-\dfrac12\right)=\dfrac54C, so C=-\dfrac{24}{5}.

Integrating: \displaystyle\int\dfrac{C}{2x+3}dx=\dfrac{C}{2}\log|2x+3|, so the last term is -\dfrac{12}{5}\log|2x+3|.

\displaystyle\int \dfrac{2x-3}{(x^2-1)(2x+3)}\,dx = \dfrac52\log|x+1| - \dfrac{1}{10}\log|x-1| - \dfrac{12}{5}\log|2x+3| + C
11

Integrate: \dfrac{5x}{(x+1)(x^2-4)}

Medium +
Solution

Factor x^2-4=(x-2)(x+2), then 5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2).

Putting x=-1: -5=A(-3)(1), so A=\dfrac53. Putting x=2: 10=B(3)(4), so B=\dfrac56. Putting x=-2: -10=C(-1)(-4), so C=-\dfrac52.

Check: A+B+C=\dfrac53+\dfrac56-\dfrac52=0, as it must be. Integrating each term gives the answer below.

\displaystyle\int \dfrac{5x}{(x+1)(x^2-4)}\,dx = \dfrac53\log|x+1| + \dfrac56\log|x-2| - \dfrac52\log|x+2| + C
12

Integrate: \dfrac{x^3+x+1}{x^2-1}

Medium +
Solution

Improper — divide: x^3+x+1 = x(x^2-1)+(2x+1), so \dfrac{x^3+x+1}{x^2-1} = x+\dfrac{2x+1}{(x-1)(x+1)}.

Decomposing the remainder as \dfrac{A}{x-1}+\dfrac{B}{x+1} gives 2x+1=A(x+1)+B(x-1). Putting x=1: 3=2A, so A=\dfrac32. Putting x=-1: -1=-2B, so B=\dfrac12.

Integrating x+\dfrac{3/2}{x-1}+\dfrac{1/2}{x+1} gives the answer below.

Alternative (skip the division): write \dfrac{x^3+x+1}{x^2-1}=x+\dfrac{A}{x-1}+\dfrac{B}{x+1}, multiply through by x^2-1 and compare coefficients: x^3+x+1=x^3+(A+B-1)x+(A-B), so A+B=2 and A-B=1, giving the same A=\dfrac32, B=\dfrac12. (The shortcut marked in some other questions does not apply here, because the numerator 2x+1 has an x in it.)

\displaystyle\int \dfrac{x^3+x+1}{x^2-1}\,dx = \dfrac{x^2}{2} + \dfrac32\log|x-1| + \dfrac12\log|x+1| + C
13

Integrate: \dfrac{2}{(1-x)(1+x^2)}

Medium +
Solution

Write \dfrac{2}{(1-x)(1+x^2)} = \dfrac{A}{1-x}+\dfrac{Bx+C}{1+x^2}, so 2=A(1+x^2)+(Bx+C)(1-x).

Putting x=1: 2=2A, so A=1. Expanding, 2=(A-B)x^2+(B-C)x+(A+C); comparing the x^2 coefficients gives B=A=1, and comparing the x coefficients gives C=B=1.

So the integrand is \dfrac{1}{1-x}+\dfrac{x+1}{1+x^2}=\dfrac{1}{1-x}+\dfrac12\cdot\dfrac{2x}{1+x^2}+\dfrac{1}{1+x^2}, which integrates to the answer below.

\displaystyle\int \dfrac{2}{(1-x)(1+x^2)}\,dx = -\log|1-x| + \dfrac12\log(1+x^2) + \tan^{-1}x + C
14

Integrate: \dfrac{3x-1}{(x+2)^2}

Easy +
Solution

Write \dfrac{3x-1}{(x+2)^2}=\dfrac{A}{x+2}+\dfrac{B}{(x+2)^2}, so 3x-1=A(x+2)+B: comparing the x coefficient gives A=3, and putting x=-2 gives B=-7.

Quicker: 3x-1=3(x+2)-7, so the fraction splits directly as \dfrac{3}{x+2}-\dfrac{7}{(x+2)^2}. Since \displaystyle\int-7(x+2)^{-2}dx=\dfrac{7}{x+2}, the integral is as below.

\displaystyle\int \dfrac{3x-1}{(x+2)^2}\,dx = 3\log|x+2| + \dfrac{7}{x+2} + C
15

Integrate: \dfrac{1}{x^4-1}

Hard +
Solution

Factor fully: x^4-1=(x-1)(x+1)(x^2+1).

Write \dfrac{1}{x^4-1}=\dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{Cx+D}{x^2+1}, so 1=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x^2-1).

Putting x=1: 1=4A, so A=\dfrac14. Putting x=-1: 1=-4B, so B=-\dfrac14. Comparing the x^3 coefficients: 0=A+B+C, so C=0. Comparing the constant terms: 1=A-B-D, so D=-\dfrac12.

ShortcutTreat x^2 as one block. The factors x^2-1 and x^2+1 differ by 2 and the numerator is 1, so \dfrac{1}{x^4-1}=\dfrac12\left[\dfrac{1}{x^2-1}-\dfrac{1}{x^2+1}\right]. Applying the same shortcut again, \dfrac{1}{x^2-1}=\dfrac12\left[\dfrac{1}{x-1}-\dfrac{1}{x+1}\right], so \dfrac{1}{x^4-1}=\dfrac14\left[\dfrac{1}{x-1}-\dfrac{1}{x+1}\right]-\dfrac12\cdot\dfrac{1}{x^2+1}: the same result in two lines.

\displaystyle\int \dfrac{1}{x^4-1}\,dx = \dfrac14\log\left|\dfrac{x-1}{x+1}\right| - \dfrac12\tan^{-1}x + C
16

Integrate: \dfrac{1}{x(x^n+1)}  [Hint: multiply numerator and denominator by x^{n-1} and put x^n=t]

Hard +
Solution

Multiplying by x^{n-1} top and bottom gives \dfrac{x^{n-1}}{x^n(x^n+1)}.

Put t=x^n, so x^{n-1}dx=\dfrac{dt}{n}, reducing the integral to \dfrac1n\displaystyle\int \dfrac{dt}{t(t+1)}.

ShortcutThe numerator is 1 and the factors t and t+1 differ by 1, so \dfrac{1}{t(t+1)}=\dfrac1t-\dfrac{1}{t+1} directly.

Integrating: \dfrac1n\left[\log|t|-\log|t+1|\right]=\dfrac1n\log\left|\dfrac{t}{t+1}\right|.

\displaystyle\int \dfrac{1}{x(x^n+1)}\,dx = \dfrac1n\log\left|\dfrac{x^n}{x^n+1}\right| + C
17

Integrate: \dfrac{\cos x}{(1-\sin x)(2-\sin x)}  [Hint: Put \sin x=t]

Medium +
Solution

Put t=\sin x, so dt=\cos x\,dx, reducing the integral to \displaystyle\int \dfrac{dt}{(1-t)(2-t)}.

ShortcutThe numerator is 1 and the factors 1-t and 2-t differ by 1 (the bigger one is 2-t), so \dfrac{1}{(1-t)(2-t)}=\dfrac{1}{1-t}-\dfrac{1}{2-t} directly.

Integrating: -\log|1-t|+\log|2-t|, since \displaystyle\int\dfrac{dt}{1-t}=-\log|1-t|.

\displaystyle\int \dfrac{\cos x}{(1-\sin x)(2-\sin x)}\,dx = \log\left|\dfrac{2-\sin x}{1-\sin x}\right| + C
18

Integrate: \dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}

Hard +
Solution

Put y=x^2; the fraction \dfrac{(y+1)(y+2)}{(y+3)(y+4)}=\dfrac{y^2+3y+2}{y^2+7y+12} is improper in y. Dividing, y^2+3y+2=(y^2+7y+12)-(4y+10), so the fraction equals 1-\dfrac{4y+10}{(y+3)(y+4)}.

Decompose: \dfrac{4y+10}{(y+3)(y+4)}=\dfrac{A}{y+3}+\dfrac{B}{y+4}, so 4y+10=A(y+4)+B(y+3). Putting y=-3: -2=A. Putting y=-4: -6=-B, so B=6.

Hence the fraction is 1-\left[-\dfrac{2}{y+3}+\dfrac{6}{y+4}\right]=1+\dfrac{2}{y+3}-\dfrac{6}{y+4}. Substituting back y=x^2 gives 1+\dfrac{2}{x^2+3}-\dfrac{6}{x^2+4}.

Integrating with \displaystyle\int\dfrac{dx}{x^2+a^2}=\dfrac1a\tan^{-1}\dfrac{x}{a}: the second term gives \dfrac{2}{\sqrt3}\tan^{-1}\dfrac{x}{\sqrt3} and the third gives -6\cdot\dfrac12\tan^{-1}\dfrac{x}{2}=-3\tan^{-1}\dfrac{x}{2}.

\displaystyle\int \dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx = x + \dfrac{2}{\sqrt3}\tan^{-1}\left(\dfrac{x}{\sqrt3}\right) - 3\tan^{-1}\left(\dfrac{x}{2}\right) + C
19

Integrate: \dfrac{2x}{(x^2+1)(x^2+3)}

Easy +
Solution

Put y=x^2, so dy=2x\,dx, reducing the integral to \displaystyle\int \dfrac{dy}{(y+1)(y+3)}.

ShortcutThe numerator is 1 and the factors y+1 and y+3 differ by 2, so \dfrac{1}{(y+1)(y+3)}=\dfrac12\left[\dfrac{1}{y+1}-\dfrac{1}{y+3}\right].

Integrating: \dfrac12\left[\log|y+1|-\log|y+3|\right]; substituting back y=x^2 gives the answer below.

\displaystyle\int \dfrac{2x}{(x^2+1)(x^2+3)}\,dx = \dfrac12\log\left|\dfrac{x^2+1}{x^2+3}\right| + C
20

Integrate: \dfrac{1}{x(x^4-1)}

Hard +
Solution

Factor fully: x(x^4-1)=x(x-1)(x+1)(x^2+1).

Write \dfrac{1}{x(x^4-1)}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1}+\dfrac{Dx+E}{x^2+1}. Putting x=0: A=\dfrac{1}{-1}=-1. Putting x=1: B=\dfrac{1}{(1)(2)(2)}=\dfrac14. Putting x=-1: C=\dfrac{1}{(-1)(-2)(2)}=\dfrac14.

Comparing the x^4 coefficients: 0=A+B+C+D, so D=\dfrac12. Comparing the x^3 coefficients: 0=B-C+E, so E=0.

ShortcutMultiply the numerator and denominator by x^3: \dfrac{1}{x(x^4-1)}=\dfrac{x^3}{x^4(x^4-1)}. Put t=x^4, so x^3dx=\dfrac{dt}{4} and the integral becomes \dfrac14\displaystyle\int\dfrac{dt}{t(t-1)}. The factors t-1 and t differ by 1, so \dfrac{1}{t(t-1)}=\dfrac{1}{t-1}-\dfrac1t, giving \dfrac14\log\left|\dfrac{t-1}{t}\right|=\dfrac14\log\left|\dfrac{x^4-1}{x^4}\right|.

This is the same as the answer below, because \dfrac14\log|x^4-1|=\dfrac14\log|x-1|+\dfrac14\log|x+1|+\dfrac14\log(x^2+1) and \dfrac14\log|x^4|=\log|x|.

\displaystyle\int \dfrac{1}{x(x^4-1)}\,dx = -\log|x| + \dfrac14\log|x-1| + \dfrac14\log|x+1| + \dfrac14\log(x^2+1) + C
21

Integrate: \dfrac{1}{e^x-1}  [Hint: Put e^x=t]

Medium +
Solution

Put t=e^x, so dt=e^x\,dx, i.e. dx=\dfrac{dt}{t}. The integral becomes \displaystyle\int \dfrac{1}{t-1}\cdot\dfrac{dt}{t}=\displaystyle\int \dfrac{dt}{t(t-1)}.

ShortcutThe numerator is 1 and the factors t-1 and t differ by 1, so \dfrac{1}{t(t-1)}=\dfrac{1}{t-1}-\dfrac1t directly.

Integrating: \log|t-1|-\log|t|=\log\left|\dfrac{t-1}{t}\right|.

\displaystyle\int \dfrac{1}{e^x-1}\,dx = \log\left|\dfrac{e^x-1}{e^x}\right| + C
22

MCQ. \displaystyle\int \dfrac{x\,dx}{(x-1)(x-2)} equals   (A) \log\left|\dfrac{(x-1)^2}{x-2}\right|+C   (B) \log\left|\dfrac{(x-2)^2}{x-1}\right|+C   (C) \log\left|\left(\dfrac{x-1}{x-2}\right)^2\right|+C   (D) \log|(x-1)(x-2)|+C

Easy +
Solution

Write \dfrac{x}{(x-1)(x-2)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}, so x=A(x-2)+B(x-1). Putting x=1: 1=-A, so A=-1. Putting x=2: 2=B.

So the integral is -\log|x-1|+2\log|x-2|=\log|x-2|^2-\log|x-1|=\log\left|\dfrac{(x-2)^2}{x-1}\right|, which is option (B).

Answer: (B) \log\left|\dfrac{(x-2)^2}{x-1}\right|+C
23

MCQ. \displaystyle\int \dfrac{dx}{x(x^2+1)} equals   (A) \log|x|-\dfrac12\log(x^2+1)+C   (B) \log|x|+\dfrac12\log(x^2+1)+C   (C) -\log|x|+\dfrac12\log(x^2+1)+C   (D) \dfrac12\log|x|+\log(x^2+1)+C

Easy +
Solution

Write \dfrac{1}{x(x^2+1)} = \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}.

So 1=A(x^2+1)+(Bx+C)x. Putting x=0: A=1. Comparing the x^2 coefficients: 0=A+B, so B=-1. Comparing the x coefficients: 0=C.

So the integrand is \dfrac1x-\dfrac{x}{x^2+1}. Since \displaystyle\int\dfrac{x}{x^2+1}dx=\dfrac12\log(x^2+1), the integral is as in option (A).

Answer: (A) \log|x|-\dfrac12\log(x^2+1)+C

How these solutions are worked

Every solution above uses the methods explained in the partial fractions lesson: substituting the roots to find the constants, comparing coefficients where substitution cannot isolate a constant, and dividing first for improper fractions. Wherever a question can be solved faster with the speed trick, it is marked Shortcut (Q2, Q15, Q16, Q17, Q19, Q20 and Q21).

The shortcut in words: when the numerator is a constant and the denominator is a product of two linear factors, write the constant divided by the difference of the two factors, times (1 over the smaller factor minus 1 over the bigger one): \dfrac{1}{(x+p)(x+q)}=\dfrac{1}{q-p}\left[\dfrac{1}{x+p}-\dfrac{1}{x+q}\right], where q>p.

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Common Questions

FAQs: Chapter 7 Ex 7.5

How many questions are there in Class 12 Maths Chapter 7 Ex 7.5?

Exercise 7.5 has 23 questions in total — 21 rational functions to be integrated using partial fractions, followed by 2 MCQs.

What technique does Chapter 7 Ex 7.5 test?

It tests integration by partial fractions — decomposing a rational function into simpler fractions (distinct linear factors, repeated linear factors, or irreducible quadratic factors) that can each be integrated using standard formulae.

Where can I find the official NCERT textbook for this exercise?

The official NCERT Class 12 Maths textbook, including Chapter 7 (Integrals) and Exercise 7.5, is available for free at ncert.nic.in.

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